Prove that: $(\int_{0}^{1/2}e^x(f(1-x) - f(x))dx)^2 \le (e-1)\int_{0}^{1}(f'(x))^2dx$.

Question

Let $f:[0,1]\to\mathbb{R}$ a differentiable function, with its derivative $f'$ continuous, so that $f(0) = f(1)$.

Prove that:

$(\int_{0}^{1/2}e^x(f(1-x) - f(x))dx)^2 \le (e-1)\int_{0}^{1}(f'(x))^2dx$.

EDIT: I realise the left member of the inequality might seem wrong, but the whole integral is raised to the power of two.

What I have tried is that

We know that

$(\int_{0}^{1/2}e^x(f(1-x) - f(x))dx)^2 = (\lvert \int_{0}^{1/2}e^x(f(1-x) - f(x))dx \lvert)^2 \le (\int_{0}^{1/2} \lvert e^x(f(1-x) - f(x)) \lvert dx)^2$.

Using the C-B-S Inequality, we obtain: $(\int_{0}^{1/2}e^x(f(1-x) - f(x))dx)^2 \le\int_{0}^{1/2}e^{2x}dx \int_{0}^{1/2} (f(1-x) - f(x))^2dx = (e-1)/2 \int_{0}^{1/2} (f(1-x) - f(x))^2dx$.

So, all there would be left to prove would be that: $\int_{0}^{1/2} (f(1-x) - f(x))^2dx \le 2\int_{0}^{1/2}(f'(x))^2dx$.

I know that the function meets all requirements in order for Rolle's Theorem to be applied, so there exists $c \in (0,1)$ so that $f'(c) = 0$, but I don't know how to use that in this case.

Thank you!!

Answer 1

We have $$I:=\int\limits_0^{1/2}e^x[f(1-x)-f(x)]\,dx =\int\limits_0^{1/2}e^x\left (\int\limits_x^{1-x}f'(y)\,dy\right )\ dx\\ =\int\limits_0^{1}f'(y)\left (\int\limits_0^{\min(y,1-y)}e^x\,dx\right )\,dy=\int\limits_0^{1}f'(y)[e^{\min(y,1-y)}-1]\,dy$$ As $\min(y,1-y)\le {1\over 2}$ we get $$|I|\le (e^{1/2}-1)\int\limits_0^{1}|f'(y)|\,dy$$ By the Cauchy-Schwarz inequality we obtain $$I^2\le (e^{1/2}-1)^2\int\limits_0^{1}f'(y)^2\,dy$$ We have $$(e^{1/2}-1)^2<(e^{1/2}-1)(e^{1/2}+1)=e-1$$ the desired inequality follows.